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Prove that $(n-r)!(r!)$ divides $ n! $ i know its a factorial formula and it might be easy but i stuck .I tried induction to $n$ or analyzing the factorials but im missing something

Jam
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2 Answers2

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You can use a combinatorial argument. How many groups of size $k$ can be formed from $k$ students? There are $n$ options for the first student, $n-1$ options for the second $\dots (n-k+1)$ options for the $k$'th student. Therefore there are $n\cdot (n-1)\cdot\dots (n-k+1)=\frac{n!}{(n-k)!}$ possible groups.

However we have overcounted! Because in a group of students there is no such thing as a first student or a second student. Every group of student has been counted exactly $k!$ times since this is the number of ways to order the $k$ students of a given group. Therefore the number of groups is $\frac{n!}{k!(n-k)!}$. Clearly the number of groups is an integer number. Then $\frac{n!}{k!(n-k)!}$ is an integer and so $k!(n-k)!$ divides $n!$

Asinomás
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    I don't see why the number of groups must be an integer. Why can't you cut a student in half? – Emily Jan 25 '15 at 17:50
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    Cutting a student in half yields at most one student if you are very lucky. In most cases the result is zero students – Asinomás Jan 25 '15 at 17:52
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Here is another solution via group theory. Consider the group $S_n$, it has $n!$ elements. Now consider the subgroup of $S_n$ consisting of the permutations such that elements in $\{1,2,3\dots k\}$ always go to elements of $\{1,2,3\dots k\}$, it has order $k!(n-k)!$ so by lagrange's theorem $k!(n-k)!$ divides $n!$

Asinomás
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  • why is that the number of the order of the subgroup? – Jam Jan 25 '15 at 18:04
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    There are $k!$ ways to permute the elements of ${1,2,3,4\dots k}$ and there are $n-k$ ways to permute the elements of ${k+1,k+2\dots n}$ – Asinomás Jan 25 '15 at 18:06